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Optical Receivers 199

Example 5.3
Consider radiation of wavelength = 700 nm incident on a photodetector whose measured responsivity is
0.4 A/W. What is its quantum efficiency at this wavelength? If the wavelength is reduced to 500 nm, what is
the new QE assuming that the responsivity is the same?

Solution:
Using Eq. (5.18), we get
(μm) 0.4 × 1.24
0
0.4A/W = ⇒ = = 0.7086(≃ 71%).
1.24 0.7
For = 500 nm, the new QE is
0
0.5
= 0.7086 × = 0.506(≃ 51%).
0.7

5.2.3 Photodetector Design Rules

As shown in Eq. (5.15), to improve the quantum efficiency, we should minimize light reflections (R term)
p
from the semiconductor surface or maximize the light transmitted into the semiconductor. For this, we can
use an antireflection coating to achieve better light transmittance. If the light is incident from air (refractive
index n ) into the semiconductor (refractive index n ), then we should choose a material whose refractive
air
sc
index n AR (refractive index of antireflection coating) is given by
√
n AR = n n . (5.19)
air sc
If we use a quarter-wavelength antireflection coating of a transparent material with a refractive index n ,
AR
then the thickness t which causes minimum reflection of the incoming radiation is given by (see Section
AR
6.6.3)

t AR = , (5.20)
4n AR
where is the free-space wavelength of the incident light onto the antireflection coating.

Example 5.4
If we use a silicon photodetector to detect red light at 680 nm, and refractive index of air (n )= 1, refractive
air
index of silicon (n )= 3.6, determine the refractive index and thickness of the antireflection coating.
Si
Solution:
√
The required antireflection coating should have a refractive index n AR = n n = 1.9 and its thickness
air Si
should be t AR = ∕(4n AR )= 680 nm∕(4 × 1.9)≃ 90 nm. At 680 nm, the refractive index of silicon nitride
(Si N )∼ 2 and that of silicon dioxide (SiO )∼ 1.5. Therefore, Si N would be a good (though not perfect)
4
3
2
4
3